CCEA A2 Biology Formula & Theorem Quick-Reference Handbook | CCEA A2 生物公式定理速查手册

📚 CCEA A2 Biology Formula & Theorem Quick-Reference Handbook | CCEA A2 生物公式定理速查手册

This quick-reference handbook brings together the essential formulae, theorems, and key quantitative concepts required for the CCEA Year 13 Biology specification (A2 1: Physiology, Co-ordination and Control, and Ecosystems). Each entry is presented with a clear statement of the principle, the standard equation, a worked example, and a note on its application within the syllabus. The aim is to provide a reliable revision tool that allows you to check definitions, carry out calculations with confidence, and interpret results in an examination context.

本速查手册汇总了 CCEA Year 13 生物学（A2 1：生理学、协调控制与生态系统）所需的核心公式、定理和关键量化概念。每条内容都包含原理说明、标准方程式、示例计算以及在大纲中的应用提示，旨在提供一份可靠的复习工具，帮助你在考试中准确进行运算并理解结果的意义。

1. Hardy–Weinberg Principle | 哈代–温伯格原理

The Hardy–Weinberg principle states that in a large, randomly mating population where no mutation, migration, or natural selection occurs, allele and genotype frequencies remain constant from generation to generation. The two equations are used to calculate the frequency of alleles and genotypes in a population.

哈代–温伯格原理指出，在一个足够大、随机交配且没有突变、迁移和自然选择的群体中，等位基因频率和基因型频率将世代保持恒定。两个方程用于计算群体中的等位基因频率和基因型频率。

The allele frequency equation is p + q = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. The genotype frequency equation is p² + 2pq + q² = 1, where p² represents the frequency of homozygous dominant individuals, 2pq the frequency of heterozygotes, and q² the frequency of homozygous recessive individuals.

等位基因频率方程为 p + q = 1，其中 p 为显性等位基因频率，q 为隐性等位基因频率。基因型频率方程为 p² + 2pq + q² = 1，p² 代表显性纯合子频率，2pq 代表杂合子频率，q² 代表隐性纯合子频率。

Example: In a population of 500 pea plants, 20 are dwarf (recessive phenotype). Therefore q² = 20/500 = 0.04, so q = √0.04 = 0.2. Since p + q = 1, p = 0.8. The frequency of heterozygotes is 2pq = 2 × 0.8 × 0.2 = 0.32, meaning 32% of the population are carriers.

示例：在一个 500 株豌豆的群体中，20 株为矮茎（隐性性状）。因此 q² = 20/500 = 0.04，q = √0.04 = 0.2。由 p + q = 1 得 p = 0.8。杂合子频率 2pq = 2 × 0.8 × 0.2 = 0.32，即 32% 的个体为携带者。

Exam tip: Always begin by identifying the homozygous recessive frequency from the question, convert it to q², and solve for q before calculating 2pq.

考试提示：务必从题中找出隐性纯合子的频率，转换为 q² 并求出 q，再计算 2pq。

2. Chi-Squared (χ²) Test | 卡方（χ²）检验

The chi-squared test is a statistical tool used to determine whether there is a significant difference between observed and expected frequencies in categorical data. In biology, it is frequently applied to genetics crosses and ecological distribution studies.

卡方检验用于判断分类数据中观察值与预期值之间是否存在显著差异。在生物学中，常应用于遗传杂交和生态分布研究。

The formula is χ² = Σ (O − E)² / E, where O is the observed value and E is the expected value. The sum is taken over all categories. The calculated χ² value is compared against a critical value from the chi-squared distribution table at the chosen probability level (usually p = 0.05) and degrees of freedom (number of categories minus one).

计算公式为 χ² = Σ (O − E)² / E，其中 O 为观察值，E 为预期值。对所有类别求和。计算出的 χ² 值与卡方分布表中选定概率水平（通常 p = 0.05）和自由度（类别数减 1）下的临界值进行比较。

If χ² calculated < χ² critical, the null hypothesis is accepted: any difference is due to chance. If χ² calculated ≥ χ² critical, the null hypothesis is rejected: there is a significant difference.

若 χ² 计算值 < χ² 临界值，则接受原假设：差异由偶然造成。若 χ² 计算值 ≥ χ² 临界值，则拒绝原假设：存在显著差异。

Worked example: In a dihybrid cross, expected phenotypes follow a 9:3:3:1 ratio. Observed counts are 90, 28, 32, 10. Expected counts based on total 160 are 90, 30, 30, 10. χ² = (0)²/90 + (−2)²/30 + (2)²/30 + (0)²/10 = 0 + 0.133 + 0.133 + 0 = 0.266. With degrees of freedom 3, the critical value at p=0.05 is 7.815. Since 0.266 < 7.815, the observed data fit the expected ratio.

示例：双因子杂交预期表型比为 9:3:3:1。观察值分别为 90、28、32、10（总数 160）。预期值为 90、30、30、10。χ² = (0)²/90 + (−2)²/30 + (2)²/30 + (0)²/10 = 0 + 0.133 + 0.133 + 0 = 0.266。自由度 3，p=0.05 时临界值 7.815。由于 0.266 < 7.815，观察数据符合预期比例。

3. Standard Deviation | 标准差

Standard deviation measures the spread of data around the mean. It is particularly useful for comparing the variation within two or more sets of biological measurements, such as lengths of leaves or enzyme reaction rates, and is essential when interpreting error bars on graphs.

标准差衡量数据围绕平均值的分散程度，特别适用于比较两组或多组生物测量数据的变异大小，如叶片长度或酶反应速率，并在解释图表中的误差线时必不可少。

The formula is s = √[ Σ(x − x̄)² / (n − 1) ], where x represents each individual value, x̄ is the mean, and n is the sample size. The denominator (n−1) gives the sample standard deviation, which is an unbiased estimator for the population standard deviation.

公式为 s = √[ Σ(x − x̄)² / (n − 1) ]，x 为单个数据值，x̄ 为平均值，n 为样本量。分母 (n−1) 用于计算样本标准差，是总体标准差的无偏估计量。

Example: For five seedlings with heights (cm) 4, 5, 7, 8, 6, the mean x̄ = 6. The deviations are −2, −1, 1, 2, 0. Σ(x − x̄)² = 4+1+1+4+0 = 10. n−1 = 4, so s = √(10/4) = √2.5 ≈ 1.58 cm. A larger standard deviation indicates greater variation from the mean.

示例：5 株幼苗高度（cm）为 4、5、7、8、6，均值 x̄ = 6。偏差为 −2、−1、1、2、0。Σ(x − x̄)² = 4+1+1+4+0 = 10。n−1 = 4，s = √(10/4) = √2.5 ≈ 1.58 cm。标准差越大，表示数据围绕均值的变异越大。

Context: When two means are compared, overlapping standard deviation error bars often suggest that the difference may not be statistically significant.

情景：比较两个均值时，标准差异差线重叠往往表明差异可能不具有统计学显著性。

4. Net Primary Production (NPP) | 净初级生产力（NPP）

Net primary production is the energy that remains in primary producers after accounting for their own respiratory losses. It represents the energy available to the next trophic level in a food chain. The relationship is given by NPP = GPP − R, where GPP is gross primary production (total energy fixed by photosynthesis) and R is respiratory loss.

净初级生产力指初级生产者扣除自身呼吸消耗后剩余的能量，代表了食物链中可供下一营养级利用的能量。关系式为 NPP = GPP − R，其中 GPP 为总初级生产力（光合作用固定的总能量），R 为呼吸消耗。

Units are typically kilojoules per square metre per year (kJ m⁻² yr⁻¹) or grams of biomass per square metre per year (g m⁻² yr⁻¹). GPP, NPP, and R must all be expressed in the same units for direct calculation.

单位通常为千焦每平方米每年（kJ m⁻² yr⁻¹）或克生物量每平方米每年（g m⁻² yr⁻¹）。GPP、NPP 和 R 必须采用相同单位才能直接计算。

Example: A forest ecosystem has a GPP of 30,000 kJ m⁻² yr⁻¹ and plant respiration accounts for 18,000 kJ m⁻² yr⁻¹. NPP = 30,000 − 18,000 = 12,000 kJ m⁻² yr⁻¹. This energy is then available to herbivores, although not all will be assimilated.

示例：某森林生态系统 GPP 为 30,000 kJ m⁻² yr⁻¹，植物呼吸消耗 18,000 kJ m⁻² yr⁻¹。NPP = 30,000 − 18,000 = 12,000 kJ m⁻² yr⁻¹。这部分能量可供植食动物利用，尽管并非全部被同化。

CCEA note: Questions frequently ask for the calculation of NPP or the percentage of solar energy converted to GPP/NPP, and may link to pyramids of energy or productivity.

CCEA 提示：考题常要求计算 NPP，或计算太阳能转化为 GPP/NPP 的百分比，并可能与能量金字塔或生产力相关联。

5. Respiratory Quotient (RQ) | 呼吸商（RQ）

The respiratory quotient is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed during respiration. It provides information about the respiratory substrate being metabolised.

呼吸商是呼吸过程中产生的二氧化碳体积与消耗的氧气体积之比，可提供被代谢的呼吸底物类型的信息。

The formula is RQ = CO₂ produced / O₂ consumed. Typical RQ values are 1.0 for carbohydrates, approximately 0.7 for lipids, and around 0.8–0.9 for proteins, though proteins are less easily characterised due to the variable handling of nitrogenous waste.

公式为 RQ = 产生的 CO₂ / 消耗的 O₂。碳水化合物的典型 RQ 值为 1.0，脂类约为 0.7，蛋白质约为 0.8–0.9，但因含氮废物处理方式不同，蛋白质的 RQ 较难表征。

Example: A respirometer experiment shows that a germinating seed absorbs 25 cm³ of O₂ and releases 25 cm³ of CO₂ in the same period. RQ = 25/25 = 1.0, indicating that carbohydrates are the main respiratory substrate. When RQ falls below 1.0, lipid or protein metabolism is suggested.

示例：呼吸计实验显示萌发种子在相同时间内吸收 25 cm³ O₂ 并释放 25 cm³ CO₂。RQ = 25/25 = 1.0，表明主要呼吸底物为碳水化合物。当 RQ 低于 1.0 时，提示脂质或蛋白质代谢。

Practical application: RQ determination using a simple respirometer (with soda lime or potassium hydroxide to absorb CO₂) is a core practical skill. The change in gas volume can be measured with a manometer, and RQ calculated under aerobic conditions.

实际应用：使用简单呼吸计（用碱石灰或氢氧化钾吸收 CO₂）测定 RQ 是一项核心实验技能。气体体积变化可通过检压计测量，并在有氧条件下计算 RQ。

6. Population Growth Rate | 种群增长率

Population growth rate expresses the change in population size over a given time interval, accounting for births, deaths, immigration, and emigration. The basic formula is: population growth rate = (births − deaths) + (immigration − emigration), often expressed per unit time and per capita for comparative purposes.

种群增长率表示在特定时间间隔内种群数量的变化，涉及出生、死亡、迁入和迁出。基本公式为：种群增长率 =（出生数 − 死亡数）+（迁入数 − 迁出数），常以每单位时间或人均形式表达以便比较。

The per capita growth rate (r) is given by r = (birth rate − death rate) / initial population size, assuming a closed population. If births exceed deaths, r is positive and the population grows; if deaths exceed births, r is negative.

人均增长率（r）表示为 r =（出生率 − 死亡率）/ 初始种群数量，假设为封闭种群。若出生大于死亡，r 为正，种群增长；若死亡大于出生，r 为负。

Example: A population of 2000 rabbits produces 500 births and 300 deaths in one year. r = (500−300)/2000 = 0.1 per year. The population will increase by 10% per year if no other factors intervene.

示例：某 2000 只兔的种群一年内出生 500 只，死亡 300 只。r = (500−300)/2000 = 0.1 每年。若无其他因素干扰，种群每年增长 10%。

In exponential growth, population size Nₜ = N₀eʳᵗ, where N₀ is the initial population size, e is Euler’s number, r is the intrinsic rate of increase, and t is time. This model assumes unlimited resources.

在指数增长中，种群数量 Nₜ = N₀eʳᵗ，其中 N₀ 为初始种群数量，e 为自然常数，r 为内禀增长率，t 为时间。该模型假设资源无限。

7. Logistic Growth Equation | 逻辑斯蒂增长方程

Logistic growth describes population growth that slows as it approaches the carrying capacity (K) of the environment. The rate equation for logistic growth is dN/dt = rN (K − N)/K, where N is population size, r is the per capita growth rate, and K is the carrying capacity.

逻辑斯蒂增长描述种群增长随接近环境容纳量（K）而减缓的过程。逻辑斯蒂增长的速率方程为 dN/dt = rN (K − N)/K，其中 N 为种群数量，r 为人均增长率，K 为环境容纳量。

When N is small, (K−N)/K ≈ 1, and growth approximates exponential. As N increases, (K−N)/K decreases, reducing the growth rate until N = K, where growth stops. The graph is S-shaped (sigmoid).

当 N 很小时，(K−N)/K ≈ 1，增长近似于指数增长。随着 N 增大，(K−N)/K 减小，增长率下降，直到 N = K 时增长停止，曲线呈 S 形（S 型）。

Example: A bacterial culture has r = 0.5 h⁻¹ and K = 10⁶ cells mL⁻¹. When N = 2×10⁵, dN/dt = 0.5 × 2×10⁵ × (10⁶ − 2×10⁵)/10⁶ = 10⁵ × 0.8 = 8×10⁴ cells mL⁻¹ h⁻¹. The rate declines as N approaches K.

示例：某细菌培养物 r = 0.5 h⁻¹，K = 10⁶ cells mL⁻¹。当 N = 2×10⁵ 时，dN/dt = 0.5 × 2×10⁵ × (10⁶ − 2×10⁵)/10⁶ = 10⁵ × 0.8 = 8×10⁴ cells mL⁻¹ h⁻¹。随着 N 趋近 K，速率下降。

This model applies to many natural populations and is essential for understanding sustainable yields in conservation and resource management.

该模型适用于许多自然种群，对理解保护生物学和资源管理中的可持续产量至关重要。

8. Lincoln Index (Capture-Mark-Recapture) | 林肯指数（标记重捕法）

The Lincoln index is used to estimate the size of a mobile animal population. Animals are captured, marked, and released; later a second sample is taken. The proportion of marked individuals in the second sample is used to estimate total population size N.

林肯指数用于估计活动动物种群的大小。先捕获动物并标记后释放，之后再采集第二个样本，利用第二个样本中标记个体的比例来估算总种群数 N。

The formula is N = (M × C) / R, where M is the number of individuals marked and released in the first sample, C is the total number captured in the second sample, and R is the number of marked individuals recaptured in the second sample.

公式为 N = (M × C) / R，其中 M 为第一次捕获并标记的个体数，C 为第二次捕获的总个体数，R 为第二次捕获中已标记的个体数。

Example: In a hedgehog survey, 30 individuals are caught, marked, and released (M=30). Two nights later, 50 hedgehogs are caught (C=50), of which 15 are marked (R=15). Estimated population size N = (30 × 50) / 15 = 1500 / 15 = 100.

示例：在某刺猬调查中，首次捕获标记并释放 30 只（M=30）。两夜后捕获 50 只（C=50），其中 15 只有标记（R=15）。估算种群数量 N = (30 × 50) / 15 = 1500 / 15 = 100 只。

Assumptions include: the marked individuals mix randomly with the population, marking does not affect survival, there is no immigration or emigration, and marks are not lost between sampling events. Violating these assumptions can lead to over- or underestimates.

该方法假设：标记个体与种群随机混合、标记不影响生存、无迁入迁出、两次采样间标记不会丢失。违反这些假设可能导致高估或低估。

9. Simpson’s Diversity Index | 辛普森多样性指数

Simpson’s Diversity Index (D) measures the biodiversity of a habitat, taking into account both species richness and the evenness of abundance across species. The formula is D = 1 − Σ (n/N)², where n is the number of individuals of a particular species and N is the total number of individuals of all species.

辛普森多样性指数（D）衡量栖息地的生物多样性，同时考虑物种丰富度和各物种个体数的均匀度。公式为 D = 1 − Σ (n/N)²，n 为某一物种的个体数，N 为所有物种个体总数。

A high value of D (close to 1) indicates high diversity; a low value (close to 0) indicates low diversity and high dominance by a few species. Some textbooks use the reciprocal form 1/Σ(n/N)², but the CCEA specification commonly uses D = 1 − Σ(n/N)².

D 值高（接近 1）表示多样性高；D 值低（接近 0）表示多样性低，少数物种占优势。某些教材使用倒数形式 1/Σ(n/N)²，但 CCEA 大纲通常使用 D = 1 − Σ(n/N)²。

Example: In a pond, there are 40 water boatmen, 30 pond skaters, 20 newts, and 10 beetles. N = 100. Σ(n/N)² = (0.4)²+(0.3)²+(0.2)²+(0.1)² = 0.16+0.09+0.04+0.01 = 0.30. D = 1 − 0.30 = 0.70, suggesting a fairly diverse community.

示例：某池塘有 40 只划蝽、30 只水黾、20 只蝾螈和 10 只甲虫。N = 100。Σ(n/N)² = (0.4)²+(0.3)²+(0.2)²+(0.1)² = 0.16+0.09+0.04+0.01 = 0.30。D = 1 − 0.30 = 0.70，表明群落多样性较高。

This index is commonly used in fieldwork to compare two habitats or assess the impact of environmental change on biodiversity.

该指数常用于野外工作中比较两个栖息地或评估环境变化对生物多样性的影响。

10. Surface Area to Volume Ratio | 表面积与体积之比

The surface area to volume ratio (SA:V) is a fundamental concept in biology that explains why larger organisms require specialised exchange surfaces and transport systems. As an organism increases in size, its volume grows faster than its surface area, so the ratio decreases.

表面积与体积比（SA:V）是生物学中的一个基本概念，可解释为什么较大生物体需要特化的交换表面和运输系统。随体型增大，体积增长快于表面积，因此比值下降。

For a sphere, surface area = 4πr² and volume = 4/3 πr³, so SA:V = 3/r. For a cube of side L, SA = 6L², V = L³, so SA:V = 6/L. Though no single formula is tested in isolation, the principle is applied to heat exchange, gas diffusion, and nutrient absorption.

对球体，表面积 = 4πr²，体积 = 4/3 πr³，因此 SA:V = 3/r。对边长为 L 的立方体，SA = 6L²，V = L³，SA:V = 6/L。尽管不单独考查某一公式，但相关原理应用于热交换、气体扩散和营养吸收。

Example: A small cube of side 1 cm has SA:V = 6:1; a larger cube of side 3 cm has SA:V = 54:27 = 2:1. The smaller cube has a ratio three times larger, allowing more efficient exchange per unit volume.

示例：边长 1 cm 的小立方体 SA:V = 6:1；边长 3 cm 的大立方体 SA:V = 54:27 = 2:1。小立方体的比值是大立方体的 3 倍，单位体积的交换效率更高。

In exam answers, you should link declining SA:V to adaptations such as alveoli, villi, and branched circulatory systems that increase the effective surface area.

在答案中，应联系 SA:V 下降去解释肺泡、绒毛和分支循环系统等增大有效表面积的适应。

11. Magnification Formula | 放大倍数公式

Magnification is a fundamental skill in microscopy, biometric measurement, and the interpretation of photomicrographs and drawings. The formula is: Magnification = Image size / Actual object size. This relationship can be rearranged depending on which quantity is unknown.

放大倍数是显微镜操作、生物测量以及解读显微照片和绘图中的基本技能。公式为：放大倍数 = 图像尺寸 / 实际物体尺寸。可根据未知量变换算式。

Units are critical: image size and actual size must be in the same units. Typically, image size is measured in millimetres, converted to micrometres (×1000), and actual size is expressed in micrometres (µm). Remember: 1 mm = 1000 µm.

单位至关重要：图像尺寸与实际尺寸必须采用相同单位。通常图像尺寸以毫米测量，转换为微米（×1000），实际尺寸以微米表示。记住 1 mm = 1000 µm。

Example: A cell diagram has a labelled length of 60 mm. The magnification is given as ×4000. Actual length = 60 mm / 4000 = 0.015 mm = 15 µm. Conversely, if a cell of diameter 7 µm is drawn to a width of 28 mm, magnification = 28 × 1000 / 7 = 4000.

示例：一个细胞图示标注长度为 60 mm，放大倍数为×4000。实际长度 = 60 mm / 4000 = 0.015 mm = 15 µm。反之，若直径为 7 µm 的细胞画成宽度 28 mm，放大倍数 = 28 × 1000 / 7 = 4000。

Always check whether the question asks for magnification, image size, or actual size, and show all unit conversions clearly.

务必先明确题目要求的是放大倍数、图像尺寸还是实际尺寸，并清晰展示所有单位转换。

12. Rate of Reaction (Enzyme Activity) | 反应速率（酶活性）

Rate of reaction in enzyme-controlled experiments is often calculated as Rate = 1 / time taken to reach a measurable end point, or Rate = change in product (or substrate) concentration / time. For example, if a catalase experiment uses hydrogen peroxide, the volume of oxygen produced per unit time gives the rate.

酶控实验中的反应速率常计算为速率 = 1 / 达到可测终点所需的时间，或速率 = 产物（或底物）浓度变化 / 时间。例如，使用过氧化氢的过氧化氢酶实验，以单位时间产生的氧气体积表示速率。

The initial rate of reaction is the preferred measure to avoid complications from substrate depletion or product inhibition. It is calculated from the tangent to the early, linear part of a progress curve, giving Rate = Δ[Product] / Δt or ΔVolume / Δt.

初始反应速率是避免底物耗尽或产物抑制影响的首选度量，通过绘制反应进程曲线早期线性部分的切线求得：速率 = Δ[产物]/Δt 或 Δ体积/Δt。

Example: In the first 30 seconds, 12 cm³ of oxygen is produced. Initial rate = 12 cm³ / 0.5 min = 24 cm³ min⁻¹. Plotting rate against enzyme concentration, substrate concentration, or temperature allows analysis of kinetic effects.

示例：前 30 秒产生 12 cm³ 氧气。初始速率 = 12 cm³ / 0.5 min = 24 cm³ min⁻¹。以速率对酶浓度、底物浓度或温度作图，可分析动力学效应。

When temperature coefficients (Q₁₀) are required, remember Q₁₀ = rate at (T+10)°C / rate at T°C, a value typically around 2 for enzyme-catalysed reactions below the optimum temperature.

若涉及温度系数（Q₁₀），Q₁₀ = 在 (T+10)°C 下的速率 / 在 T°C 下的速率，在低于最适温度时，酶促反应的 Q₁₀ 通常约为 2。

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CCEA A2 Biology Formula & Theorem Quick-Reference Handbook | CCEA A2 生物公式定理速查手册